## anonymous 5 years ago limit of 3x^3+ln(x)/x+3x^2 as x approaches infinity.

1. amistre64

maybe a salnt asymp?

2. anonymous

? $\frac{3x^3+ln(x)}{x+3x^2}$

3. amistre64

slant even

4. anonymous

saint asymptote, the patron saint of division

5. amistre64

otherwise i think its just inf

6. anonymous

$3x^3+\ln(x)/x+2x^3$ typo'd sorry, dunno if that makes a difference.

7. amistre64

y = 3x might be the slant..

8. amistre64

it does lol

9. amistre64

3/2

10. anonymous

I'm bad at limits. Haha any help is greatly appreciated.

11. amistre64

the infinity limits are gona be at the horizontal asymps if they exist

12. anonymous

still infinity

13. amistre64

i think the ln(x)/x^3 goes to zero...

14. anonymous

oh i see it is $\frac{3x^3+ln(x)}{x+2x^3}$ my mistake i misread it. amistre right, 3/2

15. amistre64

http://www.wolframalpha.com/input/?i=lim {x+to+infinity}lnx%2Fx^3 im right lol

16. anonymous

yes you are right of course. ration of leading coefficients.

17. anonymous

ratio too

18. amistre64

ration? now you are even typing like me ;)

19. anonymous

How did you find that? How would you cancel out the natural log? I'm still kinda confused...

20. amistre64

the log tends to slow down at inifinty; so the x^3 takes over and goes to 0

21. anonymous

Ahh and then ratio of coefficients. Got it. Thanks!

22. amistre64

a the ends; log(x) is pretty much a straight line

23. anonymous

log grows slower than ANY polynomial, so you can ignore it safely

24. anonymous

Right, right, I'm in compsci. We covered that. Should have remembered. Thanks so much! This was the first question I asked, how do I close this question?

25. amistre64

you cant, its exists now forever in the ether of the cosmos ;)

26. anonymous

That's... Unfortunate.

27. anonymous

lord i hope not with all my wrong answers!

28. amistre64

if you swear alot they might delete it lol